# Tetration Forum

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(08/31/2010, 06:51 AM)mike3 Wrote: [ -> ]The product of exp is the exponential of a formal power series. This can be expressed using the Bell polynomials:

$\prod_{n=0}^{\infty} \exp(a_n y^n) = \exp\left(\sum_{n=0}^{\infty} a_n y^n\right) = \exp(a_0) \exp\left(\sum_{n=1}^{\infty} a_n y^n\right)$.

This then becomes

$\exp(a_0) \exp\left(\sum_{n=1}^{\infty} a_n y^n\right) = \exp(a_0) \sum_{n=1}^{\infty} \frac{\sum_{k=1}^n B_{n,k}(1! a_1, ..., (n-k+1)! a_{n-k+1})}{n!} y^n = \exp(a_0) \sum_{n=1}^{\infty} \frac{B_n(1! a_1, ..., n! a_n)}{n!} y^n$.

Thus the equations to solve are

$a_n L^n = \exp(a_0) \frac{B_n(1! a_1, ..., n! a_n)}{n!}$.

Since $a_0 = L$ and $a_1 = 1$, this is

$a_n L^n = L \frac{B_n(1, 2! a_2, ..., n! a_n)}{n!}$.

This is derived from Faà di Bruno's formula, see

http://en.wikipedia.org/wiki/Fa%C3%A0_di...7s_formula

for details.

Doing some tests, it appears that

$B_n(1, 2! a_2, ..., n! a_n)$

has only one occurrence of $n! a_n$, and no higher powers of it, and it never seems to be multiplied by any sort of n-dependent coefficient. This means that $B_n(1, a_2, ..., a_n) - n! a_n = B_n(1, 2! a_2, ..., (n-1)! a_{n-1}, 0)$. I don't have a proof at this point, but we can then solve this:

$a_n n! L^n = L B_n(1, 2! a_2, ..., n! a_n)$

$a_n n! L^{n-1} = B_n(1, 2! a_2, ..., n! a_n)$

$a_n n! L^{n-1} - n! a_n = B_n(1, 2! a_2, 3! a_3, ..., (n-1)! a_{n-1}, 0)$

$n! (L^{n-1} - 1) a_n = B_n(1, 2! a_2, 3! a_3, ..., (n-1)! a_{n-1}, 0)$

$a_n = \frac{B_n(1, 2! a_2, 3! a_3, ..., (n-1)! a_{n-1}, 0)}{n! (L^{n-1} - 1)}$.

And this is the recurrent formula for the general coefficients. Together with $a_0 = L$ and $a_1 = 1$, this completes the as-close-to-explicit-as-possible-so-far formula for the coefficients of the Fourier series for the regular iteration.

EDIT: posts corrected to include factorials on terms $a_n$ in Bell polynomials
(08/31/2010, 07:08 AM)Gottfried Wrote: [ -> ]Hi Sheldon -

Gottfried, thanks for your reply. I'm trying to make sense of the coefficients, and would probably need to program the matrix into pari-gp to verify it.

Earlier, Gottfried wrote:
Quote:Hi Sheldon -

just to allow me to follow (think I can't involve much) - I don't have a clue from where this is coming, what, for instance, is L at all? I think you've explained it elsewhere before but don't see it at the moment... Would you mind to reexplain in short or to provide the link?

Gottfried
L is the fixed point of base(e). If you iterate the natural logarithm function hundreds of times (say starting with z=0.5), you get a very good approximation of L. Then the equations describe the regular superfunction for base e, which is complex valued at the real axis, but is analytic and entire.

The rest of what I did basically, it comes down to expressing the regular superfunction as an analytic Fourier series, where all of the terms of the series decay to zero at i*infinity. Such a series is guaranteed to be analytic.

Here is an example of a generalized view of such a Fourier series, not related to the superfunction, with a period of 2Pi.
$f(z) = \sum_{n=0}^{\infty}a_ne^{i*nz}$
A full Fourier series would also have the negative coefficients as well, and is often not analytic.
$f(z) = \sum_{n=-\infty}^{\infty}a_ne^{i*nz}$
Both of these can be wrapped around the unit circle, with the substitution $y=e^{iz}$. In the analytic case, we have an analytic Taylor series. $f(z) = \sum_{n=0}^{\infty}a_ny^n$. In the general case, with terms from -infinty to +infinty, we have a Laurent series, with singularities inside the unit circle, and an annular ring of convergence. For the full Fourier series, often, the Laurent series only converges on the edge of the unit circle. Hope that helps.
- Sheldon

Mike,

Thanks a lot for your links and your post! I'll have to program it into pari-gp, and verify that it matches the discreet Fourier series terms. Next goal; rewrite the equations for other bases (except $\eta$), especially for bases<sqrt(2).
- Sheldon

(08/31/2010, 06:51 AM)mike3 Wrote: [ -> ]....
Doing some tests, it appears that

$B_n(1, 2! a_2, ..., n! a_n)$

has only one occurrence of $n! a_n$, and no higher powers of it....
$n! (L^{n-1} - 1) a_n = B_n(1, 2! a_2, 3! a_3, ..., (n-1)! a_{n-1}, 0)$

$a_n = \frac{B_n(1, 2! a_2, 3! a_3, ..., (n-1)! a_{n-1}, 0)}{n! (L^{n-1} - 1)}$.

And this is the recurrent formula for the general coefficients. Together with $a_0 = L$ and $a_1 = 1$, this completes the as-close-to-explicit-as-possible-so-far formula for the coefficients of the Fourier series for the regular iteration.

EDIT: posts corrected to include factorials on terms $a_n$ in Bell polynomials

On to the generic base implementation, for the periodic regular superfunction. Right now, I don't have a working routine for either Mike's suggestion of Bell's polynomial, or Gottfried's matrix routine. So, I'm still doing this "longhand". B is the base. L is the fixed point for base B. I started with my last post, and edited for a generic base.

$\text{superf}_{B}(z) = \lim_{n \to \infty} B^{[n](L + L\ln(B)^{z-n})} =
\lim_{n \to \infty} \exp^{[n]} ((L+(L\ln(B))^{z-n})/\ln(B))$

$\text{period}=2\pi i/\ln(L*\ln(B))$
$\text{period}=2\pi i/(L*\ln(B)+\ln(\ln(B)))$
:: verified if B=sqrt(2), L=4, period = 19.236*I

$\text{RegularSuperf_B}= \sum_{n=0}^{\infty}a_n(L\ln(B))^{nz}$
a_0=L, a_1=1,
substitute: $y=(L\ln(B))^z$

$\text{RegularSuperf_B}= \sum_{n=0}^{\infty}a_ny^n$

3) figure out the coefficients for z+1. $y_{z+1}=(L\ln(B))^{z+1}=(L\ln(B))*(L\ln(B))^z=(L\ln(B))*y$
$\text{RegularSuperf_B}(z+1)= \sum_{n=0}^{\infty}a_n(L\ln(B))^ny^n$

4) Apply the definition of the regular superfunction, f(z+1)=B^(f(z))
$\text{RegularSuperf_B}(z+1)=\exp({\text{ln(B)*RegularSuperf}(z)})$
$\text{RegularSuperf_B}(z+1) = \sum_{n=0}^{\infty}a_n(L\ln(B))^ny^n =
\exp({\sum_{n=0}^{\infty}\ln(B)a_ny^n})$

5) express this as a product. Then expand the Taylor series for each term in the product, and substitute in $a_0=L$ $a_1=1$.

$\sum_{n=0}^{\infty}a_n(L*\ln(B))^ny^n= \prod_{n=0}^{\infty}\exp({\ln(B)a_ny^n})$
$\sum_{n=0}^{\infty}a_n(L*\ln(B))^ny^n= \prod_{n=0}^{\infty}( { \sum_{p=0}^{\infty}({(\ln(B)a_ny^n)^p/p!})} ) =$
$
\\(L)\times
\\(1+\ln(B)y + \ln(B)^2y^2/2! + \ln(B)^3y^3/3! + \ln(B)^4y^4/4! + \ln(B)^5y^5/5! ....)\times
\\(1+\ln(B)a_2y^2+\ln(B)^2a_2^2y^4/2!+\ln(B)^3a_2^3y^6/3!+\ln(B)^8a_2^4y^8/4! ...)\times
\\(1+\ln(B)a_3y^3+\ln(B)^2a_3^2y^6/2!+\ln(B)^3a_3^3y^9/3!+ ...)\times
\\(1+\ln(B)a_4y^4+\ln(B)^2a_4^2y^8/2!+\ln(B)^3a_4^3y^{12}/3!+ ...)\times
\\(1+\ln(B)a_5y^5+\ln(B)^2a_5^2y^{10}/2!+ ...)\times ...$

6) equate terms with the same y coefficient
$\ln(B)^2a_2L^2y^2=\ln(B)^2Ly^2/2! + \ln(B)La_2y^2$
$\ln(B)^3a_3L^3y^3=\ln(B)^3Ly^3/3! + \ln(B)La_3y^3 + \ln(B)^2La_2y^3$
$\ln(B)^4a_4L^4y^4=\ln(B)^4Ly^4/4! + \ln(B)La_4y^4 + \ln(B)^3La_2y^4/2 + \ln(B)^2La_3y^4 + \ln(B)^2La_2^2y^4/2$
$\ln(B)^5a_5L^5y^5=\ln(B)^5Ly^5/5! + \ln(B)La_5y^5 + \ln(B)^4La_2y^5/6 + \ln(B)^2La_4y^5 + \ln(B)^3La_2^2y^5/2 + \ln(B)^3La_3y^5/2 + \ln(B)^2La_3a_2y^5$

cancelling, and rearranging terms
$\ln(B)a_2L - a_2=\ln(B)/2!$
$\ln(B)^2a_3L^2 - a_3=\ln(B)^2/3! + \ln(B)a_2$
$\ln(B)^3a_4L^3 - a_4=\ln(B)^3/4! + \ln(B)^2a_2/2 + \ln(B)a_3 + \ln(B)a_2^2/2$
$\ln(B)^4a_5L^4 - a_5=\ln(B)^4/5! + \ln(B)^3a_2/6 + \ln(B)a_4 + \ln(B)^2a_2^2/2 + \ln(B)^2a_3/2 + \ln(B)a_3a_2$

$a_2 = (\ln(B)/2) / (\ln(B)L-1)$
$a_3 = (\ln(B)^2/3! + \ln(B)a_2) / (\ln(B)^2L^2 - 1)$
$a_4 = (\ln(B)^3/4! + \ln(B)^2a_2/2 + \ln(B)a_3 + \ln(B)a_2^2/2) / (\ln(B)^3L^3 - 1)$
$a_5 = (\ln(B)^4/5! + \ln(B)^3a_2/6 + \ln(B)a_4 + \ln(B)^2a_2^2/2 + \ln(B)^2a_3/2 + \ln(B)a_3a_2) / (\ln(B)^4L^4 - 1)$

I haven't verified these equations, though the last set matched the discreet Fourier analysis for terms a_0 through a_5. In theory, using Mike's Bell expansion, or perhaps Gottfried's matrix, one could write a program that would generate the coefficients for a known interesting base, like sqrt(2). The equations should match the upper super exponential for sqrt(2).
- Sheldon
(08/31/2010, 02:37 PM)sheldonison Wrote: [ -> ]
(08/31/2010, 07:08 AM)Gottfried Wrote: [ -> ]Hi Sheldon -

Gottfried, thanks for your reply. I'm trying to make sense of the coefficients, and would probably need to program the matrix into pari-gp to verify it.

Sheldon -

here some Pari/GP-statements to arrive at the coefficients. I just took your last substitution rules in your earlier letter. The "cryptic" function-calls are in my collection of procedures - "VE" means truncation of vectors/matrices, dFac(1,6) creates a 6x6-diagonalmatrix containing the factorials r!^1
Code:
\\ extract coefficients at x into column-vector "pc" \ps 16    \\ use only series-precision 16 (coefficients) - if more then the product of the exp-taylorseries needs long time... pc = polcoeffs( exp('a1*x) * exp('a2*x^2) * exp('a3*x^3) * exp('a4*x^4) * exp('a5*x^5)) ~ ; \\ %box are meta-commands to advise PariTTY to send the output into a matrix-display-window %box >pc dFac(1,5)*Mat(VE(pc,5) )    \\ show the coefficients scaled by factorials in window "pc" \\ result: \\                                         1 \\                                        a1 \\                                 a1^2+2*a2 \\                         a1^3+6*a2*a1+6*a3 \\  a1^4+12*a2*a1^2+24*a3*a1+(12*a2^2+24*a4) \\ now begin to substitute a1 = 1 pc1=subst(VE(pc,6),'a1,a1) %box >pc dFac(1,6)*Mat(pc1 )    \\ show the coefficients scaled by factorials \\                                                   1 \\                                                   1 \\                                              2*a2+1 \\                                       6*a2+(6*a3+1) \\                     12*a2^2+12*a2+(24*a3+(24*a4+1)) a2 = 1/2/(L-1) pc2=subst(VE(pc1,6),'a2,a2) %box >pc dFac(1,4)*Mat(VE(pc2,4) ) \\ show only 4 rows                                                     1                                                     1                                               L/(L-1)                          ((6*a3+1)*L+(-6*a3+2))/(L-1) a3 = (1/6 + a2)/(L^2-1) pc3=subst(VE(pc2,6),'a3,a3) %box >pc dFac(1,4)*Mat(VE(pc3,4) ) a4=(1/24 + a3 + a2/2 + a2^2/2)/(L^3-1) pc4=subst(VE(pc3,6),'a4, a4) %box >pc dFac(1,5)*Mat(VE(pc4,5) ) a5=(1/5! + a4+a3/2+a3*a2+a2^2/2+a2/3!)/(L^4-1) pc5=subst(VE(pc4,6),'a5, a5) %box >pc dFac(1,6)*Mat(VE(pc5,6) )                                                                              1 \\                                                                             1 \\                                                                       L/(L-1) \\                                                     (L^3+2*L^2)/(L^3-L^2-L+1) \\                                 (L^6+5*L^5+6*L^4+6*L^3)/(L^6-L^5-L^4+L^2+L-1) \\  (L^10+9*L^9+24*L^8+40*L^7+46*L^6+36*L^5+24*L^4)/(L^10-L^9-L^8+2*L^5-L^2-L+1) \\ see the coefficients at powers of L in the numerators equal that matrix-entries, for instance last row(revert the order and begin at L^4: [24  36  46  40  24  9  1] \\ denominators can be factored by (L-1),(L^2-1) and so on

Thanks also for the other explanations. It surely helps to have an entry for my own understanding of the procedure once I've got the whole idea behind...

So far -

Gottfried

Formulas for the Bell polynomials can be found here:
http://en.wikipedia.org/wiki/Bell_polynomials

and here:

http://mathworld.wolfram.com/BellPolynomial.html

There's a method involving taking a matrix determinant, a convolution formula, and a sum over partitions (we are only interested in the "complete" Bell polynomials $B_n$ here).

E.g.

$(x \diam y)_n = \sum_{j=1}^{n-1} {n \choose j} x_j y_{n-j}$

then

$B_{n,k}(x_1, ..., x_{n-k+1}) = \frac{x_n^{k \diam}}{k!}$

(where $x^{k \diam}$ is like "exponentiating" the sequence with the "diamond" operator)

and

$B_n(x_1, ..., x_n) = \sum_{k=1}^n B_{n,k}(x_1, ..., x_{n-k+1})$
For the variable-base case,

$a_n (L \log(B))^n = L \frac{B_n(\log(B), \log(B) 2! a_2, ..., \log(B) n! a_n)}{n!}$

thus

$a_n L^{n-1} \log(B)^n n! = B_n(\log(B), \log(B) 2! a_2, ..., \log(B) n! a_n)$

$a_n = \frac{B_n(\log(B), \log(B) 2! a_2, ..., \log(B) (n-1)! a_{n-1}, 0)}{(L^{n-1} \log(B)^n - \log(B)) n!}$.
Now all we need is some way to express the Riemann mapping

(08/31/2010, 07:34 PM)mike3 Wrote: [ -> ]For the variable-base case,

$a_n (L \log(B))^n = L \frac{B_n(\log(B), \log(B) 2! a_2, ..., \log(B) n! a_n)}{n!}$

thus

$a_n L^{n-1} \log(B)^n n! = B_n(\log(B), \log(B) 2! a_2, ..., \log(B) n! a_n)$

$a_n = \frac{B_n(\log(B), \log(B) 2! a_2, ..., \log(B) (n-1)! a_{n-1}, 0)}{(L^{n-1} \log(B)^n - \log(B)) n!}$.
Wow! Thanks Mike! Only matrices and statistics, "n choose k" formulas is one of my weak areas, so it might be a few days before I catch up.

I'm kind of intrigued at actually having a closed form for a real valued superfunction, like sqrt(2). I'm guessing that the number of terms required for convergence gets extremely large as z increases. From the terms I computed a couple of days ago, it looks like the terms are decreasing exponentially, which means the series acts like it has a singularity. However, since the regular superfunction is entire, we must have convergence to infinity, so the terms must eventually decrease faster than exponentially.
- Sheldon
Please excuse - this post contains some <aaarrrgggh>s
But it's worth to notice how the bell-polynomials and my -for most fellows here: still cryptic - matrix-method are related. That I didn't see this earlier

-----------------

In the mathworld-wolfram-link I find a short description of the Bell-polynomials first kind, fortunately with some example.

$\sum_{k=0}^{\infty} \frac{B_k(x)}{k!}t^k = e^{(e^t-1)x}$

This expressed with my matrix-formulae is

V(t)~ * dF(-1)*S2 * V(x) = exp( (exp(t)-1)*x)

or using my standard-matrix for x->exp(x)-1 plus one intermediate step

V(t) ~ * fS2F = V(exp(t)-1)~

write "y" for "exp(t)-1"

V(y)~ * dF(-1) * V(x) = F(-1)~* dV(y)*V(x)
= F(-1)~ * V( y*x)
= exp( y*x)
= exp( (exp(t)-1)*x)

Or in one expression:

V(t) ~ * (fS2F * dF(-1)) * V(x) = exp( (exp(t)-1)*x)

(which I can recognize immediately to be correct because I'm extremely used to that notation)

or the even simpler definition of the vector of Bell-polynomials:

B(x) = S2*V(x)

So now I see at least, how the Bell-polynomials of the *first kind* are related to my matrix-lingo.
(So I should rename fS2F into "Bell" perhaps...)

----------------------------------

But then follows the Bell-polynomials of *second kind*. And there I'm lost again... No example, no redundancy... as if the reader could not do some error in parsing a complex formula...

Is there possibly meant the iteration of the Bell-polynomials (=matrix-power of S2)?
That would be simple then...

Or, wait - looking at the *subscripted* "x" I think, that they are now coefficients for some arbitrary function developed as a powerseries with reciprocal factorials, ... and then...

Well, this seems to be just the definition of what I found out myself at the very beginning of my fiddling with this subject and called it a "matrixoperator" for some function. With the additional property, that it is (lower) triangular (because of the missing x0 in the formula in mathworld), so for instance all my U_t-matrices for the decremented exponentiation (or "U-tetration to base t" in my early speak) were such collections of Bell-polynomials of second kind (but also the schröder-matrices, just all my lower-triangular matrix-operators which I worked with the last years and have the reciprocal factorial scaling <arrggh!>)

So now - if we talk about the Bell-polynomials (first or second kind) and the symbolic representation in terms of the log(L) (or log(t)) then this is just what I solved in my earlier posted link (good to know)! Here it is again: http://go.helms-net.de/math/tetdocs/APT.htm . And Faa di Bruno/Bell-polynomials and "matrixoperators" is the same and one needs only read about one side of these notations...

Amen -
<cough>
Gottfried

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